Question

It is believed that two C-12 nuclei can react in the core of a supergiant star to form Na-23 and H-1. Calculate the energy released from this reaction for each mole of hydrogen formed. The masses of C-12, Na-23, and H-1 are 12.0000 u, 22.989767 u, and 1.007825 u, respectively. 126C + 126C →2311Na + 11H 2.16 × 1015 kJ
a. 2.16 × 105 kJ
b. 2.16 × 1011 kJ
c. 2.16 × 108 kJ
d. 2.16 × 1014 kJ

Answer 1

Answer:

c. 2.16 × 10^8 kJ

Explanation:

In the given question, 2 C-12 nuclei were used for the reaction and the mass of C-12 is 12.0000 amu. Therefore, for  2 C-12 nuclei, the mass is 2*12.0000 = 24.0000 amu.

In addition, a Na-23 and a H-1 were formed in the process. The combined mass of the products is 22.989767+1.007825 = 23.997592 amu

The mass of the reactant is different from the mass of the products. The difference = 24.0000 amu - 23.997592 amu = 0.002408 amu.

Theoretically, 1 amu = 1.66054*10^-27 kg

Thus, 0.002408 amu = 0.002408*1.66054*10^-27 kg = 3.99858*10^-30 kg

This mass difference is converted to energy and its value can be calculated using:

E = mc^2 = 3.99858*10^-30 *(299792458)^2 = 3.59374*10^-13 J

Furthermore, 1 mole of hydrogen nuclei contains 6.022*10^23 particles. Thus, we have:

E = 3.59374*10^-13 * 6.022*10^23 = 2.164*10^11 J = 2.164*10^8 kJ